Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 12

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Weekly Quiz Competition

Question 1
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The de Broglie wavelength of an electron in the $$4^{th}$$ Bohr orbit is:

A
$$2\pi a_0$$

B
$$4\pi a_0$$

C
$$8\pi a_0$$

D
$$6\pi a_0$$

Solution

De-Broglie wavelength of an electron:

$$2\pi r = n\lambda$$

$$2\pi \times \dfrac{n^2}{z} a_0 = n.\lambda$$

$$2\pi \times \dfrac{4^2}{1}a_0 = 4\lambda$$

$$\lambda = 8\pi a_0$$

Hence, the correct answer is option $$C$$.
Question 2
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If the de Broglie wavelengths associated with a proton and an $$\alpha -$$ particle are equal, then the ratio of velocities of the proton and the $$\alpha -$$ particle will be:

A
$$1:4$$

B
$$1:2$$

C
$$4:1$$

D
$$2:1$$

Solution

Given de Broglie Wavelength $$ \lambda_{p} = \lambda_{\alpha}$$

So, $$\dfrac {h}{m_{p} \times v_{p}}$$ = $$\dfrac {h}{m_{\alpha} \times v_{\alpha}}$$

$$\dfrac{v_{p}} {v_{\alpha}}$$ = $$ \dfrac {m_{\alpha}}{m_{p}}$$ = $$\dfrac{4 m_{p}} {m_{p}}$$

Because Mass of $$\alpha$$ particle is 4 times mass of proton.

So 4:1.

Option C is correct answer
Question 3
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An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

A
$$\dfrac{1}{c} \left(\dfrac{E}{2m} \right)^{1/2}$$

B
$$c (2mE)^{1/2}$$

C
$$\left(\dfrac{E}{2m} \right)^{1/2}$$

D
$$\dfrac{1}{c} \left(\dfrac{2E}{m} \right)^{1/2}$$

Solution

For electron
De - Broglie wavelength $$\lambda_c = \dfrac{h}{p}$$
where p is momentum $$p = mv$$
also by energy we have $$E = \dfrac{1}{2} mv^2$$

$$\Rightarrow E = \dfrac{1}{2} \dfrac{p^2}{m}$$

$$\Rightarrow p = \sqrt{2 mE}$$

$$\therefore \lambda_c = \dfrac{h}{\sqrt{2 mE}}$$

For photon energy $$\Rightarrow E = \dfrac{hc}{\lambda}$$

$$\Rightarrow \lambda = \dfrac{h c}{E}$$

$$\therefore \dfrac{\lambda_c}{\lambda} = \dfrac{h}{\sqrt{2 mE}} \dfrac{E}{hc}$$

$$= \dfrac{1}{C} \sqrt{\dfrac{E}{2m}}$$

option (A) is correct.
Question 4
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A particle moving with kinetic energy E has de Broglie wavelength $$\lambda$$. If energy $$\Delta E$$ is added to its energy, the wavelength become $$\lambda /2$$. Value of $$\Delta E$$, is?

A
$$2E$$

B
$$4E$$

C
$$3E$$

D
$$E$$

Solution

For Partial,

$$\lambda=\dfrac{h}{\sqrt{2\times (K.E)\times m}}$$

$$\lambda \propto \dfrac{1}{\sqrt{K.E}}$$

$$\dfrac{\lambda_1}{\lambda_2}=\sqrt{\dfrac{(K.E)_2}{(K.E)_1}}=\sqrt{\left(\dfrac{E+\Delta E}{E}\right)}$$

$$\left(\dfrac{\lambda}{\lambda/2}\right)^2=\dfrac{E+\Delta E}{E}$$

$$4=1+\dfrac{\Delta E}{E}$$

$$3=\dfrac{\Delta E}{E}$$

$$\boxed{\Delta E=3E}$$

Option (C) is correct.
Question 5
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The relative uncertainty in the period of a satellite orbiting around the earth is $$10^{-2}$$. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :

A
$$3\times 10^{-2}$$

B
$$10^{-2}$$

C
$$2\times 10^{-2}$$

D
$$6\times 10^{-2}$$

Solution

Time period, $$T = 2 \pi \sqrt{\dfrac{r^3}{GM_e}}$$

Squaring om both sides, $$T^2 = \dfrac{4 (\pi)^2 r^3}{GM}$$

Taking log, $$ln(T^2)= ln(\dfrac{4 (\pi)^2 r^3}{GM})$$

$$ln(T^2) = ln(\dfrac{4 (\pi)^2 r^3}{G})-ln(M)$$

$$2 ln(T) = ln(K)-ln(M)$$

$$2 \dfrac{\Delta{T}}{T}= \dfrac{\Delta{M}}{M}$$

$$\dfrac{\Delta{M}}{M}= 2 \times 10^{-2}$$
Question 6
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An electron of mass $$m$$ and magnitude of charge $$|e|$$ initially at rest gets accelerated by a constant electric field $$E$$. The rate of change of de-Broglie wavelength of this electron at time $$t$$ ignoring relativistic effects is:

A
$$\dfrac {-h}{|e| Et^2}$$

B
$$-\dfrac {h}{|e| Et}$$

C
$$-\dfrac {h}{|e| Et \sqrt t}$$

D
$$\dfrac {|e| Et}{h}$$

Solution

$$P=\dfrac{h}{\lambda_D}$$

$$\lambda_D=\dfrac{h}{p}=\dfrac{h}{mv}$$

$$\because \dfrac{v}{t}=a\Rightarrow v=at$$

$$v=at$$

$$\because F=qE$$

$$ma=qE$$

$$\boxed{a=\dfrac{qE}{m}=\dfrac{eE}{m}}$$

So, $$v=\dfrac{|e|E}{m}t$$

$$\therefore \lambda_D=\dfrac{h}{m\left(\dfrac{|e|E}{m}\right)t}=\dfrac{h}{|e|Et}$$

Rate of change of de-Broglie wavelength is given by differentiating the wavelength w.r.t. time.
$$\boxed{\dfrac{d\lambda_D}{dt}=-\dfrac{h}{|e|Et^2}}$$
Question 7
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A proton has kinetic energy $$\mathrm{E}=100\mathrm{k}\mathrm{e}\mathrm{V}$$ which is equal to that of a photon. The wavelength of photon is $$\lambda_{2}$$ and that of proton is $$\lambda_{1}$$. The ratio of $$\lambda_{1}/\lambda_{2}$$ is proportional to

A
$$\mathrm{E}^{2}$$

B
$$\mathrm{E}^{1/2}$$

C
$$\mathrm{E}^{-1}$$

D
$$\mathrm{E}^{-1/2}$$

Solution

For proton and photon wavelength is given as -
$$\displaystyle \lambda_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}}=\frac{\mathrm{h}}{\sqrt{2\mathrm{m}\mathrm{E}}},\ \displaystyle \lambda_{\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}}=\frac{\mathrm{h}\mathrm{c}}{\mathrm{E}}$$.
So, the answer is option (B).
Question 8
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A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is :

(take the proton mass, $$m_p = (5/3) \times 10^{-27}kg$$ and

$$h/e = 4.2\times 10^{-15}J.s/C; \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 m/F; 1 fm = 10^{-15}m$$)

A
$$7\ fm$$

B
$$8\ fm$$

C
$$9\ fm$$

D
$$10\ fm$$

Solution

Given here,

$$q_{1} = 120 \times e$$

$$q_{2} = e $$ ( charge on proton)

It is based on conservation of energy of proton.

$$\dfrac{1}{2}mv^2 = k \dfrac{q_1q_2}{R}$$

$$v = \sqrt{\dfrac{2Kq_1q_2}{mR}}$$

Also, we know, de’broglie wavelength,

$$\lambda =\dfrac{h}{mv}$$

$$\lambda = \dfrac{h\sqrt{R}}{\sqrt{2mKq_1q_2}}\\$$

Thus,

$$\lambda = \dfrac{h\sqrt{10\times10^{-15}}}{\sqrt{2\times\dfrac{5}{3}\times10^{-27}\times9\times10^9\times120\times e^2}}\\$$

$$\lambda =\dfrac{\dfrac{h}{e}\times10^{-7}}{6\times10^{-8}}\\$$

$$\lambda =\dfrac{4.2\times10^{-15}\times10^{-7}}{6\times10^{-8}}\\$$

$$\lambda =7\times10^{-15}\ m = 7\ fm \\$$

Option A is correct.
Question 9
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A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30$$mW$$and the speed of light is $$3\times 10^{8}m/s$$. The final momentum of the object is:

A
$$0.3\times 10^{-17}kgms^{-1}$$

B
$$1.0\times 10^{-17}kg ms^{-1}$$

C
$$3.0\times 10^{-17}kgms^{-1}$$

D
$$9.0\times 10^{-17}kgms^{-1}$$

Solution

Sol$$. (B)$$
$$t=100\times 10^{-9} sec,\ P=30\times 10^{-3}$$Watt,
$$C=3\times 10^{8}m/s$$
Incident energy $$= Pt$$
By debroglie wavelength, $$\lambda=h/p$$
and Energy of photon is $$E=hc/\lambda$$
From above two equations, $$E=pc$$

Momentum $$p=\dfrac{Pt}{C}=\dfrac{30\times 10^{-3}\times 100\times 10^{-9}}{3\times 10^{8}}=1.0\times 10^{-17}$$ kg ms$$^{-1} $$

So, the answer is option (B).
Question 10
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Directions For Questions

When a particle is restricted to move along x-axis between $$x = 0$$ and $$x = a$$, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region,correspond to the formation of standing waves with nodes at its ends $$x=0$$ and $$x=a$$. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as $$E = p^{2}/2m$$. Thus, the energy of the particle can be denoted by a quantum number n taking values $$1, 2, 3, . $$($$n = 1$$, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line $$x = 0$$ to $$x = a$$. Take $$h = 6.6 \times 10^{-34} Js$$ and $$e = 1.6 \times 10^{-19} C.$$

...view full instructions

The speed of the particle that can take discrete values is proportional to

A
$$\mathrm{n}^{-3/2}$$

B
$$\mathrm{n}^{-1}$$

C
$$\mathrm{n}^{1/2}$$

D
$$n$$

Solution

As given in the question
Let the no. of loops in the wave be $$n$$
Let wavelength be $$\lambda$$
Debroglie wavelength
$$ \lambda = \cfrac{h}{mv}$$ ------------(1)
Also
$$ n \lambda = 2 \pi a$$ --------------(2)
Substituting above
$$ v = \cfrac{nh}{2am}$$
$$v \propto n$$
So, the answer is option (D).

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Dual Nature of Radiation and Matter Test - 12

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Dual Nature of Radiation and Matter Test - 12

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SHARING IS CARING

Question 1

$$2\pi a_0$$

$$4\pi a_0$$

$$8\pi a_0$$

$$6\pi a_0$$

Solution

Question 2

$$1:4$$

$$1:2$$

$$4:1$$

$$2:1$$

Solution

Question 3

$$\dfrac{1}{c} \left(\dfrac{E}{2m} \right)^{1/2}$$

$$c (2mE)^{1/2}$$

$$\left(\dfrac{E}{2m} \right)^{1/2}$$

$$\dfrac{1}{c} \left(\dfrac{2E}{m} \right)^{1/2}$$

Solution

Question 4

$$2E$$

$$4E$$

$$3E$$

$$E$$

Solution

Question 5

$$3\times 10^{-2}$$

$$10^{-2}$$

$$2\times 10^{-2}$$

$$6\times 10^{-2}$$

Solution

Question 6

$$\dfrac {-h}{|e| Et^2}$$

$$-\dfrac {h}{|e| Et}$$

$$-\dfrac {h}{|e| Et \sqrt t}$$

$$\dfrac {|e| Et}{h}$$

Solution

Question 7

$$\mathrm{E}^{2}$$

$$\mathrm{E}^{1/2}$$

$$\mathrm{E}^{-1}$$

$$\mathrm{E}^{-1/2}$$

Solution

Question 8

$$7\ fm$$

$$8\ fm$$

$$9\ fm$$

$$10\ fm$$

Solution

Question 9

$$0.3\times 10^{-17}kgms^{-1}$$

$$1.0\times 10^{-17}kg ms^{-1}$$

$$3.0\times 10^{-17}kgms^{-1}$$

$$9.0\times 10^{-17}kgms^{-1}$$

Solution

Question 10

$$\mathrm{n}^{-3/2}$$

$$\mathrm{n}^{-1}$$

$$\mathrm{n}^{1/2}$$

$$n$$

Solution

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